Simplify and expand the following expression: $ \dfrac{3}{r - 5}+ \dfrac{5}{3r + 24}- \dfrac{5}{r^2 + 3r - 40} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3r + 24} = \dfrac{5}{3(r + 8)}$ We can factor the quadratic in the third term: $ \dfrac{5}{r^2 + 3r - 40} = \dfrac{5}{(r - 5)(r + 8)}$ Now we have: $ \dfrac{3}{r - 5}+ \dfrac{5}{3(r + 8)}- \dfrac{5}{(r - 5)(r + 8)} $ The least common multiple of the denominators is: $ (r - 5)(r + 8)$ In order to get the first term over $(r - 5)(r + 8)$ , multiply by $\dfrac{3(r + 8)}{3(r + 8)}$ $ \dfrac{3}{r - 5} \times \dfrac{3(r + 8)}{3(r + 8)} = \dfrac{9(r + 8)}{(r - 5)(r + 8)} $ In order to get the second term over $(r - 5)(r + 8)$ , multiply by $\dfrac{r - 5}{r - 5}$ $ \dfrac{5}{3(r + 8)} \times \dfrac{r - 5}{r - 5} = \dfrac{5(r - 5)}{(r - 5)(r + 8)} $ In order to get the third term over $(r - 5)(r + 8)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{5}{(r - 5)(r + 8)} \times \dfrac{3}{3} = \dfrac{15}{(r - 5)(r + 8)} $ Now we have: $ \dfrac{9(r + 8)}{(r - 5)(r + 8)} + \dfrac{5(r - 5)}{(r - 5)(r + 8)} - \dfrac{15}{(r - 5)(r + 8)} $ $ = \dfrac{ 9(r + 8) + 5(r - 5) - 15} {(r - 5)(r + 8)} $ Expand: $ = \dfrac{9r + 72 + 5r - 25 - 15}{3r^2 + 9r - 120} $ $ = \dfrac{14r + 32}{3r^2 + 9r - 120}$